Nilpotency class three is 3-local for Lie rings

This article gives a proof/explanation of the equivalence of multiple definitions for the term Lie ring of nilpotency class three
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a Lie ring $L$ :

The nilpotency class of $L$ is at most three, i.e., we have $[w, [x, [y, z]]] = 0$ for all $w, x, y, z \in L$ (where we allow some or all of $w, x, y, z$ to be equal).
The 3-local nilpotency class of $L$ is at most three, i.e., the subring generated by any three elements of $L$ is nilpotent of class at most three.

Proof

(1) implies (2)

This is immediate.

(2) implies (1)

The proof is incomplete.

Given: A Lie ring $L$ of 3-local nilpotency class three

To prove: $[w, [x, [y, z]]] = 0$ for all $w, x, y, z \in L$

Proof:

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	The map $(w, x, y, z) \mapsto [w, [x, [y, z]]]$ is alternating and multi-linear, and hence skew-symmetric, in all pairs of inputs. In particular, this means that its sign is reversed under any odd permutation of the inputs and is preserved under any even permutation of the inputs.	3-local class three		For any pair of inputs, if we set them to be equal, then we have a situation of a Lie bracket of length four involving three elements, which must be zero. Hence, we have the alternating condition.
2	For all $w, x, y, z \in L$ , we have $3 [w, [x, [y, z]]] = 0$ .		Step (1)	By Step (1), we have $[w, [x, [y, z]]] = [w, [y, [z, x]]] = [w, [z, [x, y]]]$ . By linearity in $w$ and the Jacobi identity in $x, y, z$ , the sum is zero, hence we get the result.
3	For all $w, x, y, z \in L$ , we have $[w, [x, [y, z]]] = [[y, z], [w, x]]$		Step (2)	Use the Jacobi identity to get $[w, [x, [y, z]]] + [x, [[y, z], w]] + [[y, z], [x, w]] = 0$ . Now, the middle term $[x, [[y, z], w]]$ is the negative of $[x, [w, [y, z]]]$ , which by the alternating condition we know to be the negative of $[w, [x, [y, z]]]$ . So, the middle term equals $[w, [x, [y, z]]]$ . Thus, the Jacobi identity expression gives $2 [w, [x, [y, z]]] + [[y, z], [w, x]] = 0$ . Combine with the conclusion of Step (2) to obtain that $[w, [x, [y, z]]] = [[y, z], [w, x]]$ .
4	For all $w, x, y, z \in L$ , we have $[y, [z, [w, x]]] = [[w, x], [y, z]]$		Step (3)	Interchange the roles of $w$ with $y$ , and $x$ with $z$ , in Step (3).
5	For all $w, x, y, z \in L$ , we have $[[y, z], [w, x]] = [[w, x], [y, z]]$		Steps (1), (3), (4)	Combine Steps (3) and (4) with the observation that, by Step (1), $< m a t h > [w, [x, [y, z]]] = [y, [z, [w, x]]]$ because the underlying permutation is even.
6	For all $w, x, y, z \in L$ , we have $[[y, z], [w, x]] = - [[w, x], [y, z]]$			Use the skew symmetry of the Lie bracket
7	For all $w, x, y, z \in L$ , we have $2 [[y, z], [w, x]] = 0$		Steps (5), (6)	Step-combination direct
8	For all $w, x, y, z \in L$ , we have $2 [w, [x, [y, z]]] = 0$		Steps (3), (7)	Step-combination direct
9	For all $w, x, y, z \in L$ , we have $[w, [x, [y, z]]] = 0$		Steps (2), (8)	Step-combination direct